Let $V$ be a simple solid region oriented with outward normals that has a piecewise-smooth boundary surface $S$. $ \iiint_V y\left( x + e^x \right) \, dV$ Use the divergence theorem to rewrite the triple integral as a surface integral. Leave out constant coefficients and extraneous functions of $x$ and $y$. $ \oiint_S (ye^x \hat{\imath} + z \hat{\jmath} + $ + + + + + + ​ + + + - $\cdot$ $\frac{x}{y}$ $\sqrt{x}$ $\sqrt[3]{x}$ $a^b$ $\pi$ Sorry, I don't understand that! $ \hat{k}) \cdot dS$
Explanation: Assume we have a simple solid region $V$ oriented with outward normals, and it has a piecewise-smooth, closed boundary surface $S$. If $F$ is a continuously differentiable vector field in $\mathbb{R}^3$, then the divergence theorem says: $ \oiint_S F \cdot dS = \iiint_V \text{div}(F) \, dV$ The given surface and boundary satisfy the conditions for the divergence theorem. We're converting a triple integral into a surface integral, so we know $\text{div}(F) = y\left( x + e^x \right)$ and we want to find $F$. We already have two components: $F(x, y, z) = ye^x \hat{\imath} + z \hat{\jmath} + ( ??? ) \hat{k}$ Let's name the missing $z$ -component $P$. The only constraint we have on $P$ is that it must make $F$ have the correct divergence. $\begin{aligned} \text{div}(F) &= ye^x + \dfrac{\partial P}{\partial z} \\ \\ &= y(x + e^x) \\ \\ &= ye^x + xy \end{aligned}$ Matching terms, our original constraint now becomes one specific requirement: $\dfrac{\partial P}{\partial z} = xy$ The most general solution includes an arbitrary function of $x$ and $y$ that we can call $H(x, y)$. $P = xyz + H(x, y)$ The problem asks us to set $H(x, y) = 0$ so that we leave out extraneous functions of $x$ and $y$. Plugging $F$ into the divergence theorem, we conclude that an equivalent surface integral to the triple integral is: $ \oiint_S (ye^x \hat{\imath} + z \hat{\jmath} + xyz \hat{k}) \cdot dS$